# using trigonometry to prove that a function is less than or equal to sqrt(2)

#### hearts123

##### New member
Hey everyone! Here's a question which states that I have to solve it with trigonometry. I had the idea of substituting x as sine or cosine, because of the range but I don't know which would be more suitable. I'm supposed to substitute trigonometric identities (like sin α or something and have a range for α) to do the question. Please help asap!

f(x)= ax+b
2a^2+ 6b^2 =3
x belongs to: [-1, 1]

Prove: |f(x)| is less than or equal to sqrt(2)

#### HallsofIvy

##### Elite Member
First, divide both sides of $$\displaystyle 2a^2+ 6b^2= 3$$ by 3 to get $$\displaystyle \frac{2}{3}a^2+ 2b^2= \left(\sqrt{\frac{2}{3}}a\right)^2+ \left(\sqrt{2}b\right)^2= 1$$. Now let $$\displaystyle sin(\theta)= \sqrt{\frac{2}{3}}a$$, so that $$\displaystyle a= \sqrt{\frac{3}{2}}sin(\theta)$$ and $$\displaystyle cos(\theta)= \sqrt{2}b$$, so that $$\displaystyle b= \frac{\sqrt{2}}{2}cos(\theta)$$ (or vice-versa- it doesn't matter). What can you say about $$\displaystyle f(x)= ax+ b= \sqrt{\frac{3}{2}}sin(\theta)x+ \frac{\sqrt{2}}{2}cos(\theta)$$ with x between -1 and 1 and $$\displaystyle \theta$$ between 0 and $$\displaystyle 2\pi$$?

#### hearts123

##### New member
First, divide both sides of $$\displaystyle 2a^2+ 6b^2= 3$$ by 3 to get $$\displaystyle \frac{2}{3}a^2+ 2b^2= \left(\sqrt{\frac{2}{3}}a\right)^2+ \left(\sqrt{2}b\right)^2= 1$$. Now let $$\displaystyle sin(\theta)= \sqrt{\frac{2}{3}}a$$, so that $$\displaystyle a= \sqrt{\frac{3}{2}}sin(\theta)$$ and $$\displaystyle cos(\theta)= \sqrt{2}b$$, so that $$\displaystyle b= \frac{\sqrt{2}}{2}cos(\theta)$$ (or vice-versa- it doesn't matter). What can you say about $$\displaystyle f(x)= ax+ b= \sqrt{\frac{3}{2}}sin(\theta)x+ \frac{\sqrt{2}}{2}cos(\theta)$$ with x between -1 and 1 and $$\displaystyle \theta$$ between 0 and $$\displaystyle 2\pi$$?
Am I right in saying that the maximum possible value is sqrt(2) / 2 and minimum is - sqrt(2) / 2 ?

#### HallsofIvy

##### Elite Member
Well, how did you get that answer?

#### Jomo

##### Elite Member
Think very carefully about the statement or vice-versa- it doesn't matter that professor HallsofIvy made in a previous post. It is a very powerful statement. It should allow you to know x and $$\displaystyle \theta$$